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New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

If $y = \sqrt[]{x} + \frac{1}{\sqrt x}$ , then $\frac{d y}{d x}$ at $x = 1$ is

(a) 1

(b) $\frac{1}{2}$

(c) $\frac{1}{\sqrt 2}$

(d) 0

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t y = \sqrt[]{x} + \frac{1}{\sqrt[]{x}} \\ \frac{d y}{d x} = \frac{1}{2 \sqrt[]{x}} - \frac{1}{2 x^{3 / 2}} \\ {(\frac{d y}{d x})}_{a t x = 1} = \frac{1}{2} - \frac{1}{2} = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

0 0

New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Let $f (x) = x - [x]$ ; $x \in ℝ$

, then $f^{'} (\frac{1}{2})$ i

(a) $\frac{3}{2}$

(b) 1

(c) 0

(d) –1

0 Follower 8 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n, f (x) = x - [x] \\ W e h a v e t o f i r s t c h e c k d i f f e r e n t i a b i l i t y o f f (x) a t x = \frac{1}{2} \\ ∴ L f' (\frac{1}{2}) = L H D = \underset{h \to 0}{l i m} \frac{f [\frac{1}{2} - h] - f [\frac{1}{2}]}{- h} \\ = \underset{h \to 0}{l i m} \frac{(\frac{1}{2} - h) - [\frac{1}{2} - h] - \frac{1}{2} + [\frac{1}{2}]}{- h} = \underset{h \to 0}{l i m} \frac{\frac{1}{2} - h - 0 - \frac{1}{2} + 0}{- h} = \frac{- h}{- h} = 1 \\ L f' (\frac{1}{2}) = R H D = \underset{h \to 0}{l i m} \frac{f (\frac{1}{2} + h) - f (\frac{1}{2})}{h} \\ = \underset{h \to 0}{l i m} \frac{(\frac{1}{2} + h) - [\frac{1}{2} + h] - \frac{1}{2} + [\frac{1}{2}]}{h} = \underset{h \to 0}{l i m} \frac{\frac{1}{2} + h - 1 - \frac{1}{2} + 1}{h} = \frac{h}{h} = 1 \\ L H D = R H D \\ ∴ f' (\frac{1}{2}) = 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$l i m_{x \to 0} \frac{t a n 2 x - x}{3 x - s i n x}$ is

(a) 2

(b) $\frac{1}{2}$

(c) $- \frac{1}{2}$

(d) $\frac{1}{4}$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n, \underset{x \to 0}{l i m} \frac{t a n 2 x - x}{3 x - s i n x} \\ = \underset{x \to 0}{l i m} \frac{x [\frac{t a n 2 x}{x} - 1]}{x [3 - \frac{s i n x}{x}]} = \underset{\begin{array}{l} x \to 0 \\ ∴ 2 x \to 0 \end{array}}{l i m} \frac{\frac{t a n 2 x}{2 x} * 2 - 1}{3 - \frac{s i n x}{x}} \\ = \frac{1.2 - 1}{3 - 1} = \frac{2 - 1}{2} = \frac{1}{2} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

0 0

New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Let $f (x) = {\begin{matrix} x^{2} - 1, & 0 < x < 2, \\ 2 x + 3, & 2 \leq x < 3 \end{matrix}$ , the quadratic equation whose roots are $l i m_{x \to 2^{-}} f (x)$ and $l i m_{x \to 2^{+}} f (x)$ is

(a) $x^{2} - 6 x + 9 = 0$
(b) $x^{2} - 7 x + 8 = 0$
(c) $x^{2} - 14 x + 49 = 0$
(d) $x^{2} - 10 x + 21 = 0$

0 Follower 13 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n, f (x) = {\begin{matrix} x^{2} - 1, & 0 < x < 2 \\ 2 x + 3, & 2 \leq x < 3 \end{matrix} \\ \underset{x \to 2^{-}}{l i m} f (x) = \underset{x \to 2^{-}}{l i m} (x^{2} - 1) = \underset{h \to 0}{l i m} [{(2 - h)}^{2} - 1] = \underset{h \to 0}{l i m} (4 + h^{2} - 4 h - 1) \\ = \underset{h \to 0}{l i m} (h^{2} - 4 h + 3) = 3 \\ a n d \underset{x \to 2^{+}}{l i m} f (x) = \underset{x \to 2^{+}}{l i m} (2 x + 3) = \underset{h \to 0}{l i m} [2 (2 + h) + 3] = 7 \\ T h e r e f o r e, t h e q u a d r a t i c e q u a t i o n w h o s e r o o t s a r e 3 a n d 7 i s \\ x^{2} - (3 + 7) x + 3 * 7 = 0 i . e ., x^{2} - 10 x + 21 = 0 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

0 0

New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following
$l i m_{x \to 0} \frac{∣ s i n x ∣}{x}$ is

(a) 1

(b) –1

(c) Does not exist

(d) None of these

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n, \underset{x \to 0}{l i m} \frac{| s i n x |}{x} \\ L H L = \underset{x \to 0^{-}}{l i m} \frac{- s i n x}{x} = - 1 [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \\ R H L = \underset{x \to 0^{+}}{l i m} \frac{s i n x}{x} = 1 \\ L H L \neq R H L \\ S o, t h e limit d o e s n o t e x i s t . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

If $f (x) = {\begin{matrix} \frac{s i n [x]}{[x]}, & if [x] \neq 0, \\ 0, & [x] = 0 \end{matrix}$ , where [.] denotes the greatest integer function, then $l i m_{x \to 0} f (x)$ is equal to

(a) 1

(b) 0

(c) –1

(d) None of these

0 Follower 9 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n, f (x) = {\begin{matrix} \frac{s i n [x]}{[x]}, & i f [x] \neq 0, \\ 0, & [x] = 0 \end{matrix} \\ L H L = \underset{x \to 0^{-}}{l i m} \frac{s i n [x]}{[x]} = \underset{h \to 0}{l i m} \frac{s i n [0 - h]}{[0 - h]} = \underset{h \to 0}{l i m} \frac{- s i n [- h]}{[- h]} = - 1 \\ R H L = \underset{x \to 0^{+}}{l i m} \frac{s i n [x]}{[x]} = \underset{h \to 0}{l i m} \frac{s i n [0 + h]}{[0 + h]} = \underset{h \to 0}{l i m} \frac{s i n [h]}{[h]} = - 1 \\ L H L \neq R H L \\ S o, t h e limit d o e s n o t e x i s t . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

0 0

New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following
$\underset{x \to 1}{l i m} \frac{(\sqrt[]{x} - 1) (2 x - 3)}{2 x^{2} + x - 3}$ is

(a) $\frac{1}{10}$
(b) $- \frac{1}{10}$
(c) 1
(d) None of these

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 1}{l i m} \frac{(\sqrt[]{x} - 1) (2 x - 3)}{2 x^{2} + x - 3} \\ = \underset{x \to 1}{l i m} \frac{(\sqrt[]{x} - 1) (2 x - 3)}{2 x^{2} + 3 x - 2 x - 3} = \underset{x \to 1}{l i m} \frac{(\sqrt[]{x} - 1) (2 x - 3)}{x (2 x + 3) - 1 (2 x + 3)} \\ = \underset{x \to 1}{l i m} \frac{(\sqrt[]{x} - 1) (2 x - 3)}{(2 x + 3) (x - 1)} = \underset{x \to 1}{l i m} \frac{(\sqrt[]{x} - 1) (\sqrt[]{x} + 1) (2 x - 3)}{(2 x + 3) (x - 1) (\sqrt[]{x} + 1)} \\ = \underset{x \to 1}{l i m} \frac{(x - 1) (2 x - 3)}{(x - 1) (\sqrt[]{x} + 1) (2 x + 3)} = \underset{x \to 1}{l i m} \frac{2 x - 3}{(\sqrt[]{x} + 1) (2 x + 3)} \\ T a k i n g limits w e h a v e \\ = \frac{2 (1) - 3}{(\sqrt[]{1} + 1) (2 * 1 + 3)} = \frac{- 1}{2 * 5} = \frac{- 1}{10} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

0 0

New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$\underset{x \to \frac{π}{4}}{l i m} \frac{s e c^{2} x - 2}{t a n x - 1}$ is

(a) 3
(b) 1
(c) 0
(d) √2

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to \frac{π}{4}}{l i m} \frac{s e c^{2} x - 2}{t a n x - 1} \\ = \underset{x \to \frac{π}{4}}{l i m} \frac{1 + t a n^{2} x - 2}{t a n x - 1} = \underset{x \to \frac{π}{4}}{l i m} \frac{t a n^{2} x - 1}{t a n x - 1} = \underset{x \to \frac{π}{4}}{l i m} \frac{(t a n x + 1) (t a n x - 1)}{(t a n x - 1)} \\ = \underset{x \to \frac{π}{4}}{l i m} (t a n x + 1) = t a n \frac{π}{4} + 1 = 1 + 1 = 2 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

0 0

New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$\underset{x \to 0}{l i m} \frac{s i n x}{\sqrt[]{x + 1} - \sqrt[]{1 - x}}$ is

(a) 2
(b) 0
(c) 1
(d) -1

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{s i n x}{\sqrt[]{x + 1} - \sqrt[]{1 - x}} \\ = \underset{x \to 0}{l i m} \frac{s i n x}{\sqrt[]{x + 1} - \sqrt[]{1 - x}} * \frac{\sqrt[]{x + 1} + \sqrt[]{1 - x}}{\sqrt[]{x + 1} + \sqrt[]{1 - x}} \\ = \underset{x \to 0}{l i m} \frac{s i n x [\sqrt[]{x + 1} + \sqrt[]{1 - x}]}{x + 1 - 1 + x} = \underset{x \to 0}{l i m} \frac{s i n x [\sqrt[]{x + 1} + \sqrt[]{1 - x}]}{2 x} \\ = \frac{1}{2} . \underset{x \to 0}{l i m} \frac{s i n x}{x} [\sqrt[]{x + 1} + \sqrt[]{1 - x}] \\ T a k i n g limit, w e g e t \\ = \frac{1}{2} * 1 * [\sqrt[]{0 + 1} + \sqrt[]{1 - 0}] = \frac{1}{2} * 1 * 2 = 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

0 0

New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following
$l i m_{x \to 0} \frac{c o s e c x - c o t x}{x}$ is

(a) $- \frac{1}{2}$
(b) 1
(c) $\frac{1}{2}$
(d) 1

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{c o s e c x - c o t x}{x} \\ = \underset{x \to 0}{l i m} \frac{\frac{1}{s i n x} - \frac{c o s x}{s i n x}}{x} = \underset{x \to 0}{l i m} \frac{1 - c o s x}{x s i n x} = \underset{x \to 0}{l i m} \frac{2 s i n^{2} \frac{x}{2}}{x . 2 s i n \frac{x}{2} c o s \frac{x}{2}} \\ = \underset{x \to 0}{l i m} \frac{s i n \frac{x}{2}}{x c o s \frac{x}{2}} = \underset{x \to 0}{l i m} \frac{t a n \frac{x}{2}}{x} = \underset{x \to 0}{l i m} \frac{t a n \frac{x}{2}}{2 * \frac{x}{2}} = \frac{1}{2} * 1 = \frac{1}{2} [? \underset{x \to 0}{l i m} \frac{t a n x}{x} = 1] \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

0 0

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